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LynFran

  • one year ago

tan(1/2(cos^−1(x))=√1−x/1+x

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  1. anonymous
    • one year ago
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    Do you have any idea on starting this question?

  2. LynFran
    • one year ago
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    no is say prove that their equal

  3. anonymous
    • one year ago
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    Well that's the first step :) Do you know what to do next?

  4. LynFran
    • one year ago
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    no

  5. anonymous
    • one year ago
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    We don't need to rush in and do everything in one go :) We can firstly start of with: tan(1/2(cos^−1(x))

  6. anonymous
    • one year ago
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    Tan ( 1 ) 2 Cos^-1(x)

  7. anonymous
    • one year ago
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    Try changing the inverse trig ratio

  8. anonymous
    • one year ago
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    You get something like: Tan (Secx) 2

  9. LynFran
    • one year ago
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    ok

  10. anonymous
    • one year ago
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    So graphically, it looks something like this

  11. anonymous
    • one year ago
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    |dw:1433678743148:dw|

  12. anonymous
    • one year ago
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    The only way to get the prove this, is by using properties of Pythagoras's, since we are dealing with a right angle triangle

  13. LynFran
    • one year ago
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    ok so we finding the hypotenuse in other word?

  14. anonymous
    • one year ago
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    Yeah we could try to find the hypotenuse or the other side. Just some way to get us that Square root, to help us prove that LHS = RHS

  15. LynFran
    • one year ago
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    so we would have x^2=(1/cosx)^2+2^2)?

  16. Loser66
    • one year ago
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    Question: \(tan (\dfrac{1}{2} arccos (x))\) or \(tan (\dfrac{1}{2arccos (x))}\). which one?

  17. LynFran
    • one year ago
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    tan(12arccos(x))

  18. LynFran
    • one year ago
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    the first 1

  19. Loser66
    • one year ago
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    if so, then the right hand side must be \(\dfrac{\sqrt {1-x^2}}{1+x}\) , right?

  20. Loser66
    • one year ago
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    you missed the square at x from numerator. Am I right? please check the problem

  21. Loser66
    • one year ago
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    |dw:1433681866420:dw|

  22. LynFran
    • one year ago
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    the right side has no square at x its just; the root of (1-x)/(1+x)

  23. Loser66
    • one year ago
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    If it is so, I failed. :)

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